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Proof step 5 from proof violations-exists (previous step)

less#(a, x0) = 0, violations(x0) = 0, n < # x0, x0[n] < (a ' + x0)[n] ⊦ x0 = [], ∃ n0. x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]
Proof step 7:
less#(a, x0) = 0, violations(x0) = 0, x0[n] < (a ' + x0)[n], n < # x0 ⊦ x0 = [], ∃ n0. x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0, a ' + x0 ≠ [] ∧ n ≠ 0
Proof step 9:
less#(a, x0) = 0, violations(x0) = 0, x0[n] < (a ' + x0).tail[n - 1], n < # x0 ⊦ a ' + x0 = [], n = 0, x0 = [], ∃ n0. x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0

Rule argument

spec: list-del name: get-rec seq: ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1] subst: [n, x] → [n, a ' + x0] flags: <all paths>

Additional information

Used terms: n, a ' + x0